\(\int \frac {\sin ^7(c+d x)}{(a-b \sin ^4(c+d x))^2} \, dx\) [162]

Optimal result

Integrand size = 24, antiderivative size = 210 \[ \int \frac {\sin ^7(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\frac {\left (3 \sqrt {a}-4 \sqrt {b}\right ) \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}-\sqrt {b}}}\right )}{8 \left (\sqrt {a}-\sqrt {b}\right )^{3/2} b^{7/4} d}-\frac {\left (3 \sqrt {a}+4 \sqrt {b}\right ) \text {arctanh}\left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}+\sqrt {b}}}\right )}{8 \left (\sqrt {a}+\sqrt {b}\right )^{3/2} b^{7/4} d}-\frac {a \cos (c+d x) \left (2-\cos ^2(c+d x)\right )}{4 (a-b) b d \left (a-b+2 b \cos ^2(c+d x)-b \cos ^4(c+d x)\right )} \] Output:

1/8*(3*a^(1/2)-4*b^(1/2))*arctan(b^(1/4)*cos(d*x+c)/(a^(1/2)-b^(1/2))^(1/2 
))/(a^(1/2)-b^(1/2))^(3/2)/b^(7/4)/d-1/8*(3*a^(1/2)+4*b^(1/2))*arctanh(b^( 
1/4)*cos(d*x+c)/(a^(1/2)+b^(1/2))^(1/2))/(a^(1/2)+b^(1/2))^(3/2)/b^(7/4)/d 
-1/4*a*cos(d*x+c)*(2-cos(d*x+c)^2)/(a-b)/b/d/(a-b+2*b*cos(d*x+c)^2-b*cos(d 
*x+c)^4)
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 4.94 (sec) , antiderivative size = 565, normalized size of antiderivative = 2.69 \[ \int \frac {\sin ^7(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\frac {\frac {16 a (-5 \cos (c+d x)+\cos (3 (c+d x)))}{8 a-3 b+4 b \cos (2 (c+d x))-b \cos (4 (c+d x))}-i \text {RootSum}\left [b-4 b \text {$\#$1}^2-16 a \text {$\#$1}^4+6 b \text {$\#$1}^4-4 b \text {$\#$1}^6+b \text {$\#$1}^8\&,\frac {6 a \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right )-8 b \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right )-3 i a \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right )+4 i b \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right )-10 a \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^2+24 b \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^2+5 i a \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^2-12 i b \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^2+10 a \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^4-24 b \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^4-5 i a \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^4+12 i b \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^4-6 a \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^6+8 b \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^6+3 i a \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^6-4 i b \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^6}{-b \text {$\#$1}-8 a \text {$\#$1}^3+3 b \text {$\#$1}^3-3 b \text {$\#$1}^5+b \text {$\#$1}^7}\&\right ]}{32 (a-b) b d} \] Input:

Integrate[Sin[c + d*x]^7/(a - b*Sin[c + d*x]^4)^2,x]
 

Output:

((16*a*(-5*Cos[c + d*x] + Cos[3*(c + d*x)]))/(8*a - 3*b + 4*b*Cos[2*(c + d 
*x)] - b*Cos[4*(c + d*x)]) - I*RootSum[b - 4*b*#1^2 - 16*a*#1^4 + 6*b*#1^4 
 - 4*b*#1^6 + b*#1^8 & , (6*a*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)] - 8 
*b*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)] - (3*I)*a*Log[1 - 2*Cos[c + d* 
x]*#1 + #1^2] + (4*I)*b*Log[1 - 2*Cos[c + d*x]*#1 + #1^2] - 10*a*ArcTan[Si 
n[c + d*x]/(Cos[c + d*x] - #1)]*#1^2 + 24*b*ArcTan[Sin[c + d*x]/(Cos[c + d 
*x] - #1)]*#1^2 + (5*I)*a*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^2 - (12*I)* 
b*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^2 + 10*a*ArcTan[Sin[c + d*x]/(Cos[c 
 + d*x] - #1)]*#1^4 - 24*b*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^4 - 
 (5*I)*a*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^4 + (12*I)*b*Log[1 - 2*Cos[c 
 + d*x]*#1 + #1^2]*#1^4 - 6*a*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^ 
6 + 8*b*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^6 + (3*I)*a*Log[1 - 2* 
Cos[c + d*x]*#1 + #1^2]*#1^6 - (4*I)*b*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*# 
1^6)/(-(b*#1) - 8*a*#1^3 + 3*b*#1^3 - 3*b*#1^5 + b*#1^7) & ])/(32*(a - b)* 
b*d)
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.09, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {3042, 3694, 1517, 27, 1480, 218, 221}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\sin ^7(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (c+d x)^7}{\left (a-b \sin (c+d x)^4\right )^2}dx\)

\(\Big \downarrow \) 3694

\(\displaystyle -\frac {\int \frac {\left (1-\cos ^2(c+d x)\right )^3}{\left (-b \cos ^4(c+d x)+2 b \cos ^2(c+d x)+a-b\right )^2}d\cos (c+d x)}{d}\)

\(\Big \downarrow \) 1517

\(\displaystyle -\frac {\frac {a \cos (c+d x) \left (2-\cos ^2(c+d x)\right )}{4 b (a-b) \left (a-b \cos ^4(c+d x)+2 b \cos ^2(c+d x)-b\right )}-\frac {\int \frac {2 a \left (2 (a-2 b)-(3 a-4 b) \cos ^2(c+d x)\right )}{-b \cos ^4(c+d x)+2 b \cos ^2(c+d x)+a-b}d\cos (c+d x)}{8 a b (a-b)}}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {a \cos (c+d x) \left (2-\cos ^2(c+d x)\right )}{4 b (a-b) \left (a-b \cos ^4(c+d x)+2 b \cos ^2(c+d x)-b\right )}-\frac {\int \frac {2 (a-2 b)-(3 a-4 b) \cos ^2(c+d x)}{-b \cos ^4(c+d x)+2 b \cos ^2(c+d x)+a-b}d\cos (c+d x)}{4 b (a-b)}}{d}\)

\(\Big \downarrow \) 1480

\(\displaystyle -\frac {\frac {a \cos (c+d x) \left (2-\cos ^2(c+d x)\right )}{4 b (a-b) \left (a-b \cos ^4(c+d x)+2 b \cos ^2(c+d x)-b\right )}-\frac {-\frac {1}{2} \left (-\sqrt {a} \sqrt {b}+3 a-4 b\right ) \int \frac {1}{-b \cos ^2(c+d x)-\left (\sqrt {a}-\sqrt {b}\right ) \sqrt {b}}d\cos (c+d x)-\frac {1}{2} \left (\sqrt {a} \sqrt {b}+3 a-4 b\right ) \int \frac {1}{\left (\sqrt {a}+\sqrt {b}\right ) \sqrt {b}-b \cos ^2(c+d x)}d\cos (c+d x)}{4 b (a-b)}}{d}\)

\(\Big \downarrow \) 218

\(\displaystyle -\frac {\frac {a \cos (c+d x) \left (2-\cos ^2(c+d x)\right )}{4 b (a-b) \left (a-b \cos ^4(c+d x)+2 b \cos ^2(c+d x)-b\right )}-\frac {\frac {\left (-\sqrt {a} \sqrt {b}+3 a-4 b\right ) \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}-\sqrt {b}}}\right )}{2 b^{3/4} \sqrt {\sqrt {a}-\sqrt {b}}}-\frac {1}{2} \left (\sqrt {a} \sqrt {b}+3 a-4 b\right ) \int \frac {1}{\left (\sqrt {a}+\sqrt {b}\right ) \sqrt {b}-b \cos ^2(c+d x)}d\cos (c+d x)}{4 b (a-b)}}{d}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {\frac {a \cos (c+d x) \left (2-\cos ^2(c+d x)\right )}{4 b (a-b) \left (a-b \cos ^4(c+d x)+2 b \cos ^2(c+d x)-b\right )}-\frac {\frac {\left (-\sqrt {a} \sqrt {b}+3 a-4 b\right ) \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}-\sqrt {b}}}\right )}{2 b^{3/4} \sqrt {\sqrt {a}-\sqrt {b}}}-\frac {\left (\sqrt {a} \sqrt {b}+3 a-4 b\right ) \text {arctanh}\left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}+\sqrt {b}}}\right )}{2 b^{3/4} \sqrt {\sqrt {a}+\sqrt {b}}}}{4 b (a-b)}}{d}\)

Input:

Int[Sin[c + d*x]^7/(a - b*Sin[c + d*x]^4)^2,x]
 

Output:

-((-1/4*(((3*a - Sqrt[a]*Sqrt[b] - 4*b)*ArcTan[(b^(1/4)*Cos[c + d*x])/Sqrt 
[Sqrt[a] - Sqrt[b]]])/(2*Sqrt[Sqrt[a] - Sqrt[b]]*b^(3/4)) - ((3*a + Sqrt[a 
]*Sqrt[b] - 4*b)*ArcTanh[(b^(1/4)*Cos[c + d*x])/Sqrt[Sqrt[a] + Sqrt[b]]])/ 
(2*Sqrt[Sqrt[a] + Sqrt[b]]*b^(3/4)))/((a - b)*b) + (a*Cos[c + d*x]*(2 - Co 
s[c + d*x]^2))/(4*(a - b)*b*(a - b + 2*b*Cos[c + d*x]^2 - b*Cos[c + d*x]^4 
)))/d)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 1480

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : 
> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q))   Int[1/( 
b/2 - q/2 + c*x^2), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q))   Int[1/(b/2 
+ q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] 
 && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^2 - 4*a*c]
 

rule 1517

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x 
_Symbol] :> With[{f = Coeff[PolynomialRemainder[(d + e*x^2)^q, a + b*x^2 + 
c*x^4, x], x, 0], g = Coeff[PolynomialRemainder[(d + e*x^2)^q, a + b*x^2 + 
c*x^4, x], x, 2]}, Simp[x*(a + b*x^2 + c*x^4)^(p + 1)*((a*b*g - f*(b^2 - 2* 
a*c) - c*(b*f - 2*a*g)*x^2)/(2*a*(p + 1)*(b^2 - 4*a*c))), x] + Simp[1/(2*a* 
(p + 1)*(b^2 - 4*a*c))   Int[(a + b*x^2 + c*x^4)^(p + 1)*ExpandToSum[2*a*(p 
 + 1)*(b^2 - 4*a*c)*PolynomialQuotient[(d + e*x^2)^q, a + b*x^2 + c*x^4, x] 
 + b^2*f*(2*p + 3) - 2*a*c*f*(4*p + 5) - a*b*g + c*(4*p + 7)*(b*f - 2*a*g)* 
x^2, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ 
[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[q, 1] && LtQ[p, -1]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3694

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^ 
(p_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-ff/f 
Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - 2*b*ff^2*x^2 + b*ff^4*x^4)^p, 
 x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 
 1)/2]
 

Maple [A] (verified)

Time = 2.95 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.02

method	result	size
derivativedivides	\(\frac {\frac {\frac {a \cos \left (d x +c \right )^{3}}{4 b \left (a -b \right )}-\frac {a \cos \left (d x +c \right )}{2 b \left (a -b \right )}}{a -b +2 b \cos \left (d x +c \right )^{2}-b \cos \left (d x +c \right )^{4}}+\frac {\frac {\left (3 a \sqrt {a b}-4 \sqrt {a b}\, b -a b \right ) \arctan \left (\frac {b \cos \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-b \right ) b}}\right )}{2 \sqrt {a b}\, b \sqrt {\left (\sqrt {a b}-b \right ) b}}-\frac {\left (3 a \sqrt {a b}-4 \sqrt {a b}\, b +a b \right ) \operatorname {arctanh}\left (\frac {b \cos \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+b \right ) b}}\right )}{2 \sqrt {a b}\, b \sqrt {\left (\sqrt {a b}+b \right ) b}}}{4 a -4 b}}{d}\)	\(214\)
default	\(\frac {\frac {\frac {a \cos \left (d x +c \right )^{3}}{4 b \left (a -b \right )}-\frac {a \cos \left (d x +c \right )}{2 b \left (a -b \right )}}{a -b +2 b \cos \left (d x +c \right )^{2}-b \cos \left (d x +c \right )^{4}}+\frac {\frac {\left (3 a \sqrt {a b}-4 \sqrt {a b}\, b -a b \right ) \arctan \left (\frac {b \cos \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-b \right ) b}}\right )}{2 \sqrt {a b}\, b \sqrt {\left (\sqrt {a b}-b \right ) b}}-\frac {\left (3 a \sqrt {a b}-4 \sqrt {a b}\, b +a b \right ) \operatorname {arctanh}\left (\frac {b \cos \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+b \right ) b}}\right )}{2 \sqrt {a b}\, b \sqrt {\left (\sqrt {a b}+b \right ) b}}}{4 a -4 b}}{d}\)	\(214\)
risch	\(-\frac {a \left ({\mathrm e}^{7 i \left (d x +c \right )}-5 \,{\mathrm e}^{5 i \left (d x +c \right )}-5 \,{\mathrm e}^{3 i \left (d x +c \right )}+{\mathrm e}^{i \left (d x +c \right )}\right )}{2 b \left (a -b \right ) d \left ({\mathrm e}^{8 i \left (d x +c \right )} b -4 \,{\mathrm e}^{6 i \left (d x +c \right )} b -16 \,{\mathrm e}^{4 i \left (d x +c \right )} a +6 b \,{\mathrm e}^{4 i \left (d x +c \right )}-4 b \,{\mathrm e}^{2 i \left (d x +c \right )}+b \right )}+\frac {i \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (a^{3} b^{7} d^{4}-3 a^{2} b^{8} d^{4}+3 a \,b^{9} d^{4}-b^{10} d^{4}\right ) \textit {\_Z}^{4}+\left (-384 a^{2} b^{4} d^{2}+1920 a \,b^{5} d^{2}-2048 b^{6} d^{2}\right ) \textit {\_Z}^{2}-331776 a^{2}+1179648 a b -1048576 b^{2}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\left (\left (\frac {3 i a^{4} b^{5} d^{3}}{20736 a^{3}-103680 a^{2} b +174080 b^{2} a -98304 b^{3}}-\frac {14 i a^{3} b^{6} d^{3}}{20736 a^{3}-103680 a^{2} b +174080 b^{2} a -98304 b^{3}}+\frac {24 i a^{2} b^{7} d^{3}}{20736 a^{3}-103680 a^{2} b +174080 b^{2} a -98304 b^{3}}-\frac {18 i a \,b^{8} d^{3}}{20736 a^{3}-103680 a^{2} b +174080 b^{2} a -98304 b^{3}}+\frac {5 i b^{9} d^{3}}{20736 a^{3}-103680 a^{2} b +174080 b^{2} a -98304 b^{3}}\right ) \textit {\_R}^{3}+\left (-\frac {1728 i a^{3} b^{2} d}{20736 a^{3}-103680 a^{2} b +174080 b^{2} a -98304 b^{3}}+\frac {9856 i a^{2} b^{3} d}{20736 a^{3}-103680 a^{2} b +174080 b^{2} a -98304 b^{3}}-\frac {18368 i a \,b^{4} d}{20736 a^{3}-103680 a^{2} b +174080 b^{2} a -98304 b^{3}}+\frac {11264 i b^{5} d}{20736 a^{3}-103680 a^{2} b +174080 b^{2} a -98304 b^{3}}\right ) \textit {\_R} \right ) {\mathrm e}^{i \left (d x +c \right )}+1\right )\right )}{128}\)	\(568\)

Input:

int(sin(d*x+c)^7/(a-b*sin(d*x+c)^4)^2,x,method=_RETURNVERBOSE)
 

Output:

1/d*((1/4/b*a/(a-b)*cos(d*x+c)^3-1/2/b*a/(a-b)*cos(d*x+c))/(a-b+2*b*cos(d* 
x+c)^2-b*cos(d*x+c)^4)+1/4/(a-b)*(1/2*(3*a*(a*b)^(1/2)-4*(a*b)^(1/2)*b-a*b 
)/(a*b)^(1/2)/b/(((a*b)^(1/2)-b)*b)^(1/2)*arctan(b*cos(d*x+c)/(((a*b)^(1/2 
)-b)*b)^(1/2))-1/2*(3*a*(a*b)^(1/2)-4*(a*b)^(1/2)*b+a*b)/(a*b)^(1/2)/b/((( 
a*b)^(1/2)+b)*b)^(1/2)*arctanh(b*cos(d*x+c)/(((a*b)^(1/2)+b)*b)^(1/2))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2507 vs. \(2 (161) = 322\).

Time = 0.33 (sec) , antiderivative size = 2507, normalized size of antiderivative = 11.94 \[ \int \frac {\sin ^7(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\text {Too large to display} \] Input:

integrate(sin(d*x+c)^7/(a-b*sin(d*x+c)^4)^2,x, algorithm="fricas")
 

Output:

-1/16*(4*a*cos(d*x + c)^3 - ((a*b^2 - b^3)*d*cos(d*x + c)^4 - 2*(a*b^2 - b 
^3)*d*cos(d*x + c)^2 - (a^2*b - 2*a*b^2 + b^3)*d)*sqrt(-((a^3*b^3 - 3*a^2* 
b^4 + 3*a*b^5 - b^6)*d^2*sqrt((81*a^5 - 522*a^4*b + 1273*a^3*b^2 - 1392*a^ 
2*b^3 + 576*a*b^4)/((a^6*b^7 - 6*a^5*b^8 + 15*a^4*b^9 - 20*a^3*b^10 + 15*a 
^2*b^11 - 6*a*b^12 + b^13)*d^4)) + 3*a^2 - 15*a*b + 16*b^2)/((a^3*b^3 - 3* 
a^2*b^4 + 3*a*b^5 - b^6)*d^2))*log((81*a^3 - 405*a^2*b + 680*a*b^2 - 384*b 
^3)*cos(d*x + c) + ((3*a^4*b^5 - 14*a^3*b^6 + 24*a^2*b^7 - 18*a*b^8 + 5*b^ 
9)*d^3*sqrt((81*a^5 - 522*a^4*b + 1273*a^3*b^2 - 1392*a^2*b^3 + 576*a*b^4) 
/((a^6*b^7 - 6*a^5*b^8 + 15*a^4*b^9 - 20*a^3*b^10 + 15*a^2*b^11 - 6*a*b^12 
 + b^13)*d^4)) - 2*(9*a^3*b^2 - 47*a^2*b^3 + 82*a*b^4 - 48*b^5)*d)*sqrt(-( 
(a^3*b^3 - 3*a^2*b^4 + 3*a*b^5 - b^6)*d^2*sqrt((81*a^5 - 522*a^4*b + 1273* 
a^3*b^2 - 1392*a^2*b^3 + 576*a*b^4)/((a^6*b^7 - 6*a^5*b^8 + 15*a^4*b^9 - 2 
0*a^3*b^10 + 15*a^2*b^11 - 6*a*b^12 + b^13)*d^4)) + 3*a^2 - 15*a*b + 16*b^ 
2)/((a^3*b^3 - 3*a^2*b^4 + 3*a*b^5 - b^6)*d^2))) + ((a*b^2 - b^3)*d*cos(d* 
x + c)^4 - 2*(a*b^2 - b^3)*d*cos(d*x + c)^2 - (a^2*b - 2*a*b^2 + b^3)*d)*s 
qrt(((a^3*b^3 - 3*a^2*b^4 + 3*a*b^5 - b^6)*d^2*sqrt((81*a^5 - 522*a^4*b + 
1273*a^3*b^2 - 1392*a^2*b^3 + 576*a*b^4)/((a^6*b^7 - 6*a^5*b^8 + 15*a^4*b^ 
9 - 20*a^3*b^10 + 15*a^2*b^11 - 6*a*b^12 + b^13)*d^4)) - 3*a^2 + 15*a*b - 
16*b^2)/((a^3*b^3 - 3*a^2*b^4 + 3*a*b^5 - b^6)*d^2))*log((81*a^3 - 405*a^2 
*b + 680*a*b^2 - 384*b^3)*cos(d*x + c) + ((3*a^4*b^5 - 14*a^3*b^6 + 24*...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^7(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\text {Timed out} \] Input:

integrate(sin(d*x+c)**7/(a-b*sin(d*x+c)**4)**2,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\sin ^7(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\int { \frac {\sin \left (d x + c\right )^{7}}{{\left (b \sin \left (d x + c\right )^{4} - a\right )}^{2}} \,d x } \] Input:

integrate(sin(d*x+c)^7/(a-b*sin(d*x+c)^4)^2,x, algorithm="maxima")
 

Output:

1/2*(4*a*b*cos(2*d*x + 2*c)*cos(d*x + c) - 20*a*b*sin(3*d*x + 3*c)*sin(2*d 
*x + 2*c) + 4*a*b*sin(2*d*x + 2*c)*sin(d*x + c) - a*b*cos(d*x + c) - (a*b* 
cos(7*d*x + 7*c) - 5*a*b*cos(5*d*x + 5*c) - 5*a*b*cos(3*d*x + 3*c) + a*b*c 
os(d*x + c))*cos(8*d*x + 8*c) + (4*a*b*cos(6*d*x + 6*c) + 4*a*b*cos(2*d*x 
+ 2*c) - a*b + 2*(8*a^2 - 3*a*b)*cos(4*d*x + 4*c))*cos(7*d*x + 7*c) - 4*(5 
*a*b*cos(5*d*x + 5*c) + 5*a*b*cos(3*d*x + 3*c) - a*b*cos(d*x + c))*cos(6*d 
*x + 6*c) - 5*(4*a*b*cos(2*d*x + 2*c) - a*b + 2*(8*a^2 - 3*a*b)*cos(4*d*x 
+ 4*c))*cos(5*d*x + 5*c) - 2*(5*(8*a^2 - 3*a*b)*cos(3*d*x + 3*c) - (8*a^2 
- 3*a*b)*cos(d*x + c))*cos(4*d*x + 4*c) - 5*(4*a*b*cos(2*d*x + 2*c) - a*b) 
*cos(3*d*x + 3*c) + 2*((a*b^3 - b^4)*d*cos(8*d*x + 8*c)^2 + 16*(a*b^3 - b^ 
4)*d*cos(6*d*x + 6*c)^2 + 4*(64*a^3*b - 112*a^2*b^2 + 57*a*b^3 - 9*b^4)*d* 
cos(4*d*x + 4*c)^2 + 16*(a*b^3 - b^4)*d*cos(2*d*x + 2*c)^2 + (a*b^3 - b^4) 
*d*sin(8*d*x + 8*c)^2 + 16*(a*b^3 - b^4)*d*sin(6*d*x + 6*c)^2 + 4*(64*a^3* 
b - 112*a^2*b^2 + 57*a*b^3 - 9*b^4)*d*sin(4*d*x + 4*c)^2 + 16*(8*a^2*b^2 - 
 11*a*b^3 + 3*b^4)*d*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 16*(a*b^3 - b^4)* 
d*sin(2*d*x + 2*c)^2 - 8*(a*b^3 - b^4)*d*cos(2*d*x + 2*c) + (a*b^3 - b^4)* 
d - 2*(4*(a*b^3 - b^4)*d*cos(6*d*x + 6*c) + 2*(8*a^2*b^2 - 11*a*b^3 + 3*b^ 
4)*d*cos(4*d*x + 4*c) + 4*(a*b^3 - b^4)*d*cos(2*d*x + 2*c) - (a*b^3 - b^4) 
*d)*cos(8*d*x + 8*c) + 8*(2*(8*a^2*b^2 - 11*a*b^3 + 3*b^4)*d*cos(4*d*x + 4 
*c) + 4*(a*b^3 - b^4)*d*cos(2*d*x + 2*c) - (a*b^3 - b^4)*d)*cos(6*d*x +...
 

Giac [F]

\[ \int \frac {\sin ^7(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\int { \frac {\sin \left (d x + c\right )^{7}}{{\left (b \sin \left (d x + c\right )^{4} - a\right )}^{2}} \,d x } \] Input:

integrate(sin(d*x+c)^7/(a-b*sin(d*x+c)^4)^2,x, algorithm="giac")
 

Output:

sage0*x
 

Mupad [B] (verification not implemented)

Time = 38.46 (sec) , antiderivative size = 3612, normalized size of antiderivative = 17.20 \[ \int \frac {\sin ^7(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\text {Too large to display} \] Input:

int(sin(c + d*x)^7/(a - b*sin(c + d*x)^4)^2,x)
 

Output:

((a*cos(c + d*x)^3)/(4*b*(a - b)) - (a*cos(c + d*x))/(2*b*(a - b)))/(d*(a 
- b + 2*b*cos(c + d*x)^2 - b*cos(c + d*x)^4)) - (atan(((((1024*a*b^6 - 153 
6*a^2*b^5 + 512*a^3*b^4)/(64*(b^4 - 2*a*b^3 + a^2*b^2)) - (cos(c + d*x)*(- 
(9*a^2*(a*b^7)^(1/2) + 24*b^2*(a*b^7)^(1/2) - 15*a*b^5 + 16*b^6 + 3*a^2*b^ 
4 - 29*a*b*(a*b^7)^(1/2))/(256*(3*a*b^9 - b^10 - 3*a^2*b^8 + a^3*b^7)))^(1 
/2)*(256*a*b^6 - 512*a^2*b^5 + 256*a^3*b^4))/(4*(a^2 - 2*a*b + b^2)))*(-(9 
*a^2*(a*b^7)^(1/2) + 24*b^2*(a*b^7)^(1/2) - 15*a*b^5 + 16*b^6 + 3*a^2*b^4 
- 29*a*b*(a*b^7)^(1/2))/(256*(3*a*b^9 - b^10 - 3*a^2*b^8 + a^3*b^7)))^(1/2 
) + (cos(c + d*x)*(16*a*b^2 - 23*a^2*b + 9*a^3))/(4*(a^2 - 2*a*b + b^2)))* 
(-(9*a^2*(a*b^7)^(1/2) + 24*b^2*(a*b^7)^(1/2) - 15*a*b^5 + 16*b^6 + 3*a^2* 
b^4 - 29*a*b*(a*b^7)^(1/2))/(256*(3*a*b^9 - b^10 - 3*a^2*b^8 + a^3*b^7)))^ 
(1/2)*1i - (((1024*a*b^6 - 1536*a^2*b^5 + 512*a^3*b^4)/(64*(b^4 - 2*a*b^3 
+ a^2*b^2)) + (cos(c + d*x)*(-(9*a^2*(a*b^7)^(1/2) + 24*b^2*(a*b^7)^(1/2) 
- 15*a*b^5 + 16*b^6 + 3*a^2*b^4 - 29*a*b*(a*b^7)^(1/2))/(256*(3*a*b^9 - b^ 
10 - 3*a^2*b^8 + a^3*b^7)))^(1/2)*(256*a*b^6 - 512*a^2*b^5 + 256*a^3*b^4)) 
/(4*(a^2 - 2*a*b + b^2)))*(-(9*a^2*(a*b^7)^(1/2) + 24*b^2*(a*b^7)^(1/2) - 
15*a*b^5 + 16*b^6 + 3*a^2*b^4 - 29*a*b*(a*b^7)^(1/2))/(256*(3*a*b^9 - b^10 
 - 3*a^2*b^8 + a^3*b^7)))^(1/2) - (cos(c + d*x)*(16*a*b^2 - 23*a^2*b + 9*a 
^3))/(4*(a^2 - 2*a*b + b^2)))*(-(9*a^2*(a*b^7)^(1/2) + 24*b^2*(a*b^7)^(1/2 
) - 15*a*b^5 + 16*b^6 + 3*a^2*b^4 - 29*a*b*(a*b^7)^(1/2))/(256*(3*a*b^9...
 

Reduce [F]

\[ \int \frac {\sin ^7(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\int \frac {\sin \left (d x +c \right )^{7}}{\sin \left (d x +c \right )^{8} b^{2}-2 \sin \left (d x +c \right )^{4} a b +a^{2}}d x \] Input:

int(sin(d*x+c)^7/(a-b*sin(d*x+c)^4)^2,x)
 

Output:

int(sin(c + d*x)**7/(sin(c + d*x)**8*b**2 - 2*sin(c + d*x)**4*a*b + a**2), 
x)